Question

In an ice ring, a skater is moving at $3m{{s}^{-1}}$ and encounters a rough patch that reduces her speed by 45% , due to a friction force that is 25% of her weight. Find the length of the rough patch.
A. 1.56m
B. 1.46m
C. 1.36m
D. 1.26m

Answer 1

Hint: The work done on the skater by the frictional force is equal to change in the kinetic energy of the skater. Calculate the change in kinetic energy of the skater and equate it to the work done by the friction for the length of the rough.

Formula used:
$W=-fd$
$K=\dfrac{1}{2}m{{v}^{2}}$
$f$ is the frictional force and $d$ is the length of the rough patch .

Complete step by step answer:
When the ice skater enters the rough patch, the friction will exert a force on the skater. The direction of the frictional force is in the direction opposite to the direction of the motion of the skater. As a result, the frictional force performs a negative work on the skater and the speed of the skater reduces. The work done on a body is equal to the change in its kinetic energy i.e. $W=\Delta K$.

The work done the friction is equal to $W=-fd$,$W=-fd$, f is the frictional force and s id the length of the rough patch .It is given to 25% of the skater’s weight. The weight of the skater is equal to mg, where m is the mass of the skater and g is acceleration due to gravity. This means that $f=\dfrac{mg}{4}$.
$\Rightarrow W=-fd=-\dfrac{mgd}{4}$ …. (i)
It is said that the initial speed of the skater is ${{v}_{1}}=3m{{s}^{-1}}$ and it reduces by 45% due to the rough. This means that the final velocity is,
${{v}_{2}}={{v}_{1}}-\dfrac{45}{100}{{v}_{1}}\\
\Rightarrow{{v}_{2}} =3-\dfrac{45}{100}(3)\\
\Rightarrow{{v}_{2}} =1.65m{{s}^{-1}}$.
The kinetic energy of a body of mass m, moving with velocity v is given as $K=\dfrac{1}{2}m{{v}^{2}}$.
Therefore, the change in kinetic energy of the skater is,
$\Delta K={{K}_{2}}-{{K}_{2}}\\
\Rightarrow\Delta K=\dfrac{1}{2}mv_{1}^{2}-\dfrac{1}{2}mv_{2}^{2}$.

Substitute the values of the velocities.
$\Delta K=\dfrac{1}{2}m{{(1.65)}^{2}}-\dfrac{1}{2}m{{(3)}^{2}}$ …. (ii).
Now, equate (i) and (ii).
$\Rightarrow -\dfrac{mgd}{4}=\dfrac{1}{2}m{{(1.65)}^{2}}-\dfrac{1}{2}m{{(3)}^{2}}$
Let us considered the value of $g=10m{{s}^{-2}}$
$-\dfrac{10d}{4}=\dfrac{1}{2}\left( {{(1.65)}^{2}}-{{(3)}^{2}} \right)$
$\Rightarrow -5d=\left( 2.7225-9 \right)$
$\Rightarrow d=\dfrac{\left( 9-2.7225 \right)}{5}\\
\therefore d=1.2555\approx 1.26m$
Therefore, the length of the rough patch is 1.26m.

Hence, the correct option is D.

Note: Other than using the work energy theorem, we can also use the kinematic equations for constant acceleration.Calculate the frictional force with given information. Then we can find its acceleration and it will be constant.After this, we can the kinematic equation $2as={{v}^{2}}-{{u}^{2}}$, $a$ is its acceleration, $s$ is its displacement, $v$ is its final velocity and $u$ is the initial velocity.